Direction in proving this statement?

What does this mean, (interpretation): "As can be easily seen, only number systems
with bases 2' have an integral number of bits per symbol, and can thus be
efficiently represented in binary form?"(Adoption of the Octal Number System)

What type of proof would be feasible? Is a nonconstructive proof possible? I'm just looking for a direction, because I want to learn how to do this myself.

Again, I'm not asking anyone to do my assignment for me, just hint me at a general direction of how to prove this.

(1)
unary: digits = { 0 }, of which there is exactly one.
There is no number of bits that can hold exactly one value.
(One bit holds two values, which is too many.)

(2)
binary: digits = { 0, 1 }, of which there are exactly two.
A single bit can hold exactly two values.

(3)
ternary: digits = { 0, 1, 2 }, of which there are exactly three.
There is no number of bits that can hold exactly three values.
(One bit holds two values, which is too few; two bits holds four values, which is too many.)

(4)
quaternary: digits = { 0, 1, 2, 3 }, of which there are exactly four.
Two bits can hold exactly four values.

(5)
quinary: digits = { 0, 1, 2, 3, 4 }, of which there are exactly five.
There is no number of bits that can hold exactly five values.
(Two bits holds four values, which is too few; three bits holds eight values, which is too many.)